# 6. Comparing Alternatives

Transcription

1 IE # 6. Comparing Alternatives One of the main purposes of this course is to discuss how to make decisions in engineering economy. Let us first consider a single period case. Suppose that there are two projects: A and B. Project A requires an investment of HK$10K and will return in one year time HK$12K. The other project requires HK$100K investment and returns HK$115K next year. Clearly, irr A = 20% and irr B = 15%. Suppose that your MARR is 10%. What do you do?

2 IE First of all, we want to invest. Why? Because: P W A (MARR) = $10, $12, 000/1.1 $909 and P W B (MARR) = $100, $115, 000/1.1 $4546 Additionally, assume that: The projects are mutually exclusive. There is no limit on the budget.

3 IE According to their present value, the preference is clearly B A. According to the IRR, A looks more attractive than B. Which consideration makes more sense? Let us start from the basis that: Any investment is acceptable only when its PW under the MARR is positive.

4 IE So, we must justify the use of the additional $90,000 in Project B. In fact B \ A requires an initial investment of $90,000 and returns with $103,000. Under the MARR its present worth is $90, $103, 000/1.1 $3, 636 Conclusion: The use of additional $90,000 in Project B is justified, and so B A. Careful: The IRR can give some misleading conclusions!

5 IE To make our logic more explicit, consider the case where A will return with $20K in one year. Then, the incremental investment in B, as compared against A, is: input $(100-10)K; output $(115-20)K. Is this a good deal? (bearing in mind that we have the MARR!) Well, the PW of the incremental investment, B \ A, under the MARR 10%, is $(100 10)K + $ K = $3.6K. 1.1 So, in this case, we should not do B! Rather, we should take A.

6 IE Following this idea, we propose a general procedure for selecting one project among mutually exclusive alternatives (MEA), provided that the MARR is given. The Incremental Investment Analysis: Order the projects according to their initial capital requirement. Let be the difference of the first two remaining projects: Delete the first project if Delete the second project if IRR( ) MARR IRR( ) < MARR Repeat until there is only one project left.

7 IE Computationally speaking, the above procedure maybe cumbersome. Also, if the IRR is non-unique, then it may be difficult to implement. The problem is lessened, if we observe that for = B \ A, IRR( ) MARR only if P W B\A (MARR) = P W B (MARR) P W A (MARR) 0. This yields a simplified procedure based on the PW of the alternatives.

8 IE Projects Evaluation based on the PWs: Calculate the PWs under the MARR. Order the projects according to their initial capital requirement. Compute the PW of the difference of the first two remaining projects: Delete the first project if the PW is positive; Delete the second project if the PW is negative. Repeat until there is only one project left. In fact, one will select the one with the highest (positive) PW value.

9 IE Example: Consider projects A, B and C, all last for 10 years. The MARR is 10%. A B C initial -$390K -$920K -$660K annual $69K $167K $133.5K The order in terms of initial capital investments is: A, C and B. Let us solve the problem by the incremental IRR method. We first compare A and C by considering C \ A: Project Initial Annual C \ A -$270K $64.5K The IRR of this cash flow is 20.05% > MARR. So C A.

10 IE We delete A from further consideration, and proceed to compare C with B. The calculation shows that Project Initial Annual B \ C -$260K $33.5K IRR B\C = 3.07% < MARR. Hence we reject B, and C is selected as the result.

11 IE Now we see what happens if we apply the PW method. First of all, the PWs are computed: P W A (10%) = $390K + $69K(P/A, 10%, 10) $34K P W B (10%) = $920K + $167K(P/A, 10%, 10) $106K P W C (10%) = $660K + $133.5K(P/A, 10%, 10) $160K So, they all qualify for further consideration.

12 IE In the increasing order of initial investment: A, C and B. We consider C \ A: P W C\A 126K > 0 Therefore we take C and delete A. Now consider B \ C: P W B\C 54K < 0 So, the additional budget needed for B is worse than the MARR. Conclusion: C is most attractive under the MARR. In fact, C B A.

13 IE A mathematical question: Can the ordering produced by the incremental method be inconsistent? That is, can it be that A B and B C, but C A? The answer is: NO! Why? Because: A B iff P W A (MARR) > P W B (MARR). However, the preference relation is largely dependent on the MARR! Consider again our first example. Suppose now that the MARR is 14.8%. Then, P W A (14.8%) = / K and P W B (14.8%) = / K Hence, A becomes preferable!

14 IE Note: Of course, the evaluation based on PW is equivalent to that based on FW or AW. As we have seen in the first example, it is wrong to select the project with the highest IRR. Example: Consider 6 projects A, B, C, D, E and F. The duration for all projects are for 10 years and their cash flows are given as follows: Project Initial Investment Annual Profit Suppose that MARR = 10%. A -$900 $150 B -$1,500 $276 C -$2,500 $400 D -$4,000 $925 E -$5,000 $1,125 F -$7,000 $1,425

15 IE First of all, we can calculate the IRR of these projects: Project IRR A 10.6% B 13.0% C 9.6% D 19.1% E 18.3% F 15.6% According to the IRR, Project D turns out to be the best. But how can we interpret this fact? It is true that if the MARR is 19%, then D is the only acceptable choice. But now, the MARR is much lower, so in fact we may actually waste some investment opportunity by investing in D.

16 IE We go on carrying out the incremental analysis which results in an ordering in PWs under the current MARR. Project PW under MARR A B 196 C D E F Therefore, if we can only invest in one project, then it is E. The order is: In fact C is not qualified at all. E F D B A C

17 IE The matter will be quite different if we have a limited budget and/or the projects are not mutually exclusive. For example, suppose that we have precisely $4,000 to invest. Then, one should certainly invest in Project D. If the projects are not mutually exclusive, then the preference would be: D E F B A If IRRs are not unique, then one can use the ERR instead.

18 IE Comparing Alternatives (continue). So far, we have made the following assumptions in our analysis: The alternatives are mutually exclusive The life-span of the alternatives are the same The basic idea of the method is simple: We try to see whether any additional cent on a more expensive project is well spent, in comparison with the MARR. This is termed the incremental analysis.

19 IE In some applications indeed only one alternative is to be chosen. Example: Suppose that one must install an air compressor among 4 choices for 5 years. Their respective expenses are as follows. Project Price Annual Cost Resale Value D 1-100K -29K 10K D 2-140K -16.9K 14K D K -14.8K 25.6K D 4-122K -22.1K 14K Suppose that MARR=20%.

20 IE First of all, this is a cost alternatives, meaning that we need only to see whether the benefit from any additional cost is more attractive than the MARR or not. From D 1 to D 4, the cash flow is Price Diff Annual Saving End Value -22K 6.9K 4K Hence, IRR( D4 \D 1 ) = 20.5%. The investment is justified.

21 IE Next we consider the incremental from D 4 to D 2. Price Diff Annual Saving End Value -18K 5.2K 0K IRR( D2 \D 4 ) = 12.3%. The additional investment is NOT justified.

22 IE From D 4 to D 3. Price Diff Annual Saving End Value -26.2K 7.3K 11.6K IRR( D3 \D 4 ) = 20.4%. The additional investment is justified. Conclusion: D 3 is most attractive, provided that MARR = 20%.

23 IE We may as well use the PWs in the analysis. The PWs under the MARR are: Project D 1 D 2 D 3 D 4 PW(MARR) K K K K Therefore, the least costly alternative is D 3. It leads to the same conclusion.

24 IE Example: A downtown parking center was out of capacity. Catherine Jones, an ambitious new employee of an architectural engineering firm, was called to perform the project evaluation. Alternatives are: P : B 1 : B 2 : B 3 : Keep and improve existing parking lot Construct one-story building Construct two-story building Construct three-story building Expenses versus incomes: Project Investment Net Annual Income P -$200K $22K B 1 -$4M $600K B 2 -$5.55M $720K B 3 -$7.5M $960K

25 IE In 15 years time, the residual value of each alternative is first estimated as the same as its construction cost today. Suppose that the MARR of the firm is 10%. The IRR of each alternative is shown to be: Project IRR P 11% B 1 15% B 2 13% B % The management of the firm was tempted to choose B 1, which has the highest IRR. Catherine discovered however that the alternatives are mutually exclusive. Therefore an incremental cost analysis is more appropriate.

26 IE This boils down to calculating the PWs under the MARR: Project PWs under MARR P $15,214 B 1 $1,521,260 B 2 $1,255,062 B 3 $1,597,356 Therefore, Catherine recommended B 3. Moreover, she remarked that if the estimation of the residual values were wrong, say they were half of the construction costs, then:

27 IE Project PWs under MARR IRR P -$8, % B 1 $1,042, % B 2 $590, % B 3 $699, % In that case, she would recommend B 1. The management sees her merits and she was quickly promoted upwards in the organization.

28 IE What to do when the duration ( useful life ) of the projects are different? Obviously it is no longer fair only to compare the PWs. For example, one project, P 1, has a duration of 2 years, and the other project, P 2, has a duration of 4 years. It can be that P W P1 (MARR) < P W P2 (MARR). But, if we adopt P 1 and replace the last two years with MARR, then the combined project is more attractive than P 2.

29 IE We first consider the situation when a project can be repeated. So we can try to fill up the whole time horizon with the same repeated projects. Let the duration of P 1 be N 1, and the duration of P 2 be N 2. Let the least common multiple of N 1 and N 2 be N. Let N = n 1 N 1, and N = n 2 N 2. We then need to compare P W n1 P 1 (MARR) and P W n2 P 2 (MARR).

30 IE This is the same as to compare AW P1 (MARR) and AW P2 (MARR). Example: Consider two projects A and B with MARR=10% A B Capital Investment -$3,500 -$5,000 Annual Revenue $1,900 $2,500 Annual Expenses -$645 -$1,020 Useful life 4 years 6 years The value at the end of useful life is assumed to be zero.

31 IE Under the repeatability assumption, we need to compare 3A and 2B. Or, equivalently, their Annual Worth: The AW of A is: and the AW of B is: $3, 500(A/P, 10%, 4) + $1, 900 $645 = $151 $5, 000(A/P, 10%, 6) + $2, 500 $1, 020 = $332. Therefore, we prefer Project B. However, not all projects can be repeated. Let us consider a more general setting, when the study period is given. That is, we are given: (1) the MARR; (2) the study period N; (3) the actual project. We want to evaluate whether or not the project is more attractive than the MARR.

32 IE General Approach: Let the duration of Project P be N p. Suppose that the study period is N. Case 1. N p < N. There are two different possibilities: (A) P is a service project; (B) P is an investment project. In case of (A), one may consider repeating P till N is filled. Then truncate the possible over-time. In case of (B), one may re-invest the worth of P at the MARR till the end. Case 2. N p > N. Truncate the project at N using an estimated market value of P at N.

33 IE Consider the previous example again. Suppose the study period is changed to 6 year. Then, F W (10%) A = [ 3, 500(F/P, 10%, 4) + (1, )(F/A, 10%, 4)](F/P, 10%, 2) = $847 and F W (10%) B = 5, 000(F/P, 10%, 6) + (2, 500 1, 020)(F/A, 10%, 6) = $2, 561 So you still prefer Project B in this case.

34 IE Example: A company needs to have four additional forklift trucks to support a warehouse, which is anticipated to be shutdown in 8 years. Two mutually exclusive alternatives are identified: Truck 1 Truck 2 Capital Investment -$184K -$242K Annual Expense -$30K -$26.7K Useful life 5 years 7 years Salvage Value $17K $21K The three-year lease cost is $104K per year. The one-year lease cost is $134K per year. The MARR of the firm is 15%. Which type of the truck should the firm buy?

35 IE Let us assume that 8 is the study period. We need to compare the cash flows: Truck 1 Truck 2 Diff Year 0 -$184K -$242K -$58K Year 1 -$30K -$26.7K $3.3K Year 2 -$30K -$26.7K $3.3K Year 3 -$30K -$26.7K $3.3K Year 4 -$30K -$26.7K $3.3K Year 5 -$13K -$26.7K -$13.7K Year 6 -$104K -$26.7K $77.3K Year 7 -$104K -$5.7K $98.3K Year 8 -$104K -$134K -$30K

36 IE Since P W Diff (15%) = $5, 171, so we find Truck 2 more attractive. One may also apply the ERR method to find err Diff = 15.97%, accepting the additional price for Truck 2.

37 IE If the study period is shorter than the useful life of an equipment, then it is crucial to estimate the market value of the equipment at that time. The Imputed Market Value Technique: (implied market value) Suppose N < N p. Let MARR=r%, the initial capital investment be C and the market value at N p be S. Then, MV N = Capital Recovery (P/A, r%, N p N) +S (P/F, r%, N p N). In other words, MV N is the sum of two terms, the first being the PW at year T of the remaining Capital Recovery value, and the second is the PW at year T of the Salvage value.

38 IE Example: Suppose that we consider a project with the following data: Capital Investment = $47,600 Useful Life = 9 years Salvage Value = $5,000. Then, the capital recovery amount is $47, 600(A/P, 20%, 9) $5, 000(A/F, 20%, 9) = $11, 569. If we want to consider the end of year 5, then P W CR (20%) = $11, 569 (P/A, 20%, 4) = $29, 949 and Therefore, P W S (20%) = $5, 000(P/F, 20%, 4) = $2, 412. MV 5 = $29, $2, 412 = $32, 361.

39 IE Exercise: Consider two projects. Project 1 Project 2 Capital Investment -$40K -$60K Annual Benefit $12K $10K Useful life 4 years 8 years Salvage Value $24K $40K Suppose that MARR = 10%. We now wish to compare the projects. We assume that the projects can be repeated. Then, AW P1 (10%) = $4, 552 and AW P2 (10%) = $2, 252. So, Project 1 appears to be more attractive.

40 IE N = 8, and suppose that Project 1 cannot be repeated. In that case we will replace what remains after year 4 by MARR: F W P1 (10%) = $40K(F/P, 10%, 8) + [$12K(F/A, 10%, 4) + 24K](F/P, 10%, 4) = $30, 933 and F W P2 (10%) = $60K(F/P, 10%, 8) + $10K(F/A, 10%, 8) + $40K = $25, 743. Still Project 1 appears to be more attractive.

41 IE Assume that N = 4. We need to compute the implied market value of Project 2 at year 4. According to our formula, this value is MV 4 (P 2 ) = [60K(A/P, 10%, 8) 40K(A/F, 10%, 8)] (P/A, 10%, 4) + 40K(P/F, 10%, 4) = $51, 880. Now we can compute the Annual Worth of Project 2. AW P2 (10%, 4 years) = 60K(A/P, 10%, 4) + $10K + $51.88K(A/F, 10%, 4) = $2, 250.

42 IE What happens if the projects are not mutually exclusive? This leads to the issue of capital budgeting: maximize subject to n i=1 b ix i n i=1 c ix i C x i = 0 or 1, i = 1,..., n. Example: Budgeting with a total budget of $500,000

43 IE Project Outlay PW BC Ratio The optimal solution x = [1, 0, 1, 1, 1, 1, 0].

44 IE The budgeting problems may actually involve mutually exclusive projects. Example: Alternatives Cost PW Road Concrete, 2 lanes 2 4 Concrete, 4 lanes 3 5 Asphalt, 2 lanes Asphalt, 4 lanes Bridge Repair Add lane New Traffic Lights Lane Underpass 1 2

45 IE This results in another 0-1 integer program: max 4x 1 + 5x 2 + 3x x 4 + x x x x 8 + x 9 + 2x 10 s.t. 2x 1 + 3x x x x x x x x 9 + x 10 5 x 1 + x 2 + x 3 + x 4 1 x 5 + x 6 + x 7 1 x 8 + x 9 + x 10 1 x i = 0 or 1, i = 1, 2,..., 10. The Spreadsheet solution says that at the optimality, x 2 = 1, x 5 = 1, x 10 = 1, and all other variables are zero.

46 IE Break-even Analysis In actually decision making, it is important to take into account the uncertainty. One approach is the so-called break-even analysis. Example: Consider the cash flow Year Cash Flow 0-100M M 2 21M 3 21M 4 21M 5 21M The market value of the project at year 5 is estimated to be 51M.

47 IE But this estimation is by no means certain. What do we do? Suppose that MARR=11%. Let the market residual value be S. Then, P W = S From this formula, it is easy to see that P W 0 if and only if S 24.4M. Therefore, any residual value higher than 24.4M will make the project valuable.

48 IE Another example of this nature is as follows. Consider the following project: Investment -$1,000 Annual Return $400 Duration? MARR 10% Let the duration be N. Then, Hence P W 0 if and only if P W = 1, N 0.1. N 3.02 years.